1. Field of the Invention
The present invention relates to an integrated circuit for converting into a voltage an output current of a photoelectric conversion device such as constitutes the pickup of a player for minidiscs, compact discs, digital video disks, and the like.
2. Description of the Prior Art
In a player for such recording media as minidiscs (MDs), compact discs (CDs), and digital video disks (DVDs) that are read optically for data retrieval, a pickup is employed that is provided with a photodiode for performing photoelectric conversion and an integrated circuit (IC) for converting an output current of the photodiode into a voltage. Instead of providing the pickup with the current-to-voltage conversion IC and the photodiode as separate components, it is also possible to provide it with a photoelectric conversion IC (OEIC) that is formed by integrating those two components into a single IC.
FIG. 5 shows a circuit diagram of a conventional current-to-voltage conversion IC 20, as a first conventional example. In FIG. 5, numeral 1 represents a differential amplifier (operational amplifier), numeral 2 represents an analog switch, and numerals 51 and 52 represent resistors. The non-inverting input terminal (+) of the differential amplifier 1 is connected to a reference voltage V.sub.ref. The inverting input terminal (-) of the differential amplifier 1 is connected to an input terminal IN of the current-to-voltage conversion IC 20 as a whole, and is also connected to the output terminal of the differential amplifier 1 itself through the resistor 51 on the one hand and through the resistor 52 and the analog switch 2 that are connected in parallel with the resistor 51 on the other hand. The output terminal of the differential amplifier 1 is also connected directly to an output terminal OUT of the current-to-voltage conversion IC 20 as a whole. The analog switch 2 is turned on and off by a signal that is fed from outside to a terminal CONT.
Having the construction as described above, this current-to-voltage conversion IC 20 operates as follows. When the photodiode PD that is connected to the input terminal IN and serves as a photoelectric conversion device is not receiving light, no current flows through the feedback resistor (i.e. the resistor 51 when the analog switch 2 is off (open), and the resistors 51 and 52 as connected in parallel when the analog switch 2 is on (closed)), and therefore the voltage appearing at the output terminal OUT equals V.sub.ref. By contrast, when the photodiode PD receives light, it allows a current I to flow therethrough that is proportional to the amount of the light received. This current I flows from the output of the differential amplifier 1 through the feedback resistor, and thereby raises the voltage at the output terminal OUT by I.times.R.sub.F (R.sub.F represents the resistance of the feedback resistor). In this way, the output current of the photodiode PD is converted into a voltage.
In this current-to-voltage conversion IC 20 of the first conventional example, the following equation holds: (conversion gain)=(voltage after conversion)/(input current)=I.times.R.sub.F /I=R.sub.F. Here, if it is assumed that the resistances of the resistors 51 and 52 are R and R', respectively, the resistance R.sub.F of the feedback resistor is given as R.sub.F =R.times.R'/(R+R') when the analog switch 2 is on, and is given as R.sub.F =R when the analog switch 2 is off. This means that turning the analog switch 2 from on to off results in increasing the conversion gain.
However, the current-to-voltage conversion IC 20 of the first conventional example described above has the following disadvantage. Specifically, there, not only does the differential amplifier 1 create a pole (hereafter referred to as the "first pole") of its own, the feedback loop creates another pole (hereafter referred to as the "second pole") due to the parasitic capacitance that exists across the photodiode PD and across the feedback resistor. To obtain a desired conversion gain, the resistance R.sub.F of the feedback resistor needs to be set to a specific resistance, and this requires that the pole frequency 1/2 .pi.CR.sub.F (C represents the capacitance of the parasitic capacitance) of the second pole be so low that only a very narrow frequency range is available. In addition, as the resistance R.sub.F of the feedback resistor is increased with a view to increasing the conversion gain, the second pole shifts downward, thereby further narrowing the available frequency range.
FIG. 6 shows the frequency response of the current-to-voltage conversion IC 20. In FIG. 6, the following symbols are used:
F1P indicates the frequency of the pole created by the operational amplifier 1 itself (i.e. the first pole frequency); PA1 F2P1 indicates the frequency of the pole created by the feedback loop (i.e. the second pole frequency) when the analog switch 2 is off; PA1 F2P2 indicates the frequency of the pole created by the feedback loop (i.e. the second pole frequency) when the analog switch 2 is on; PA1 G1 indicates the loop gain (when (feedback ratio)=1); PA1 L0 indicates the open-loop gain; PA1 L1 indicates the gain obtained when the analog switch 2 is off (i.e. when the conversion gain is high); and PA1 L2 indicates the gain obtained when the analog switch 2 is on (i.e. when the conversion gain is low). PA1 F1P indicates the frequency of the pole created by the operational amplifier 1 itself (i.e. the first pole frequency); PA1 F2P indicates the frequency of the pole created by the feedback loop (i.e. the second pole frequency) (constant regardless of whether the analog switch 2 is on or off); PA1 G1 indicates the loop gain (when (feedback ratio)=R.sub.2 /(R.sub.2 +R.sub.3)); PA1 G2 indicates the loop gain (when (feedback ratio)=R.sub.2 /(R.sub.2 +R.sub.3 +R.sub.T)); PA1 L0 indicates the open-loop gain; PA1 L3 indicates the gain obtained when the analog switch 2 is on (i.e. when the conversion gain is low); and PA1 L4 indicates the gain obtained when the analog switch 2 is off (i.e. when the conversion gain is high).
FIG. 7 shows a circuit diagram of another conventional current-to-voltage conversion IC 30, as a second conventional example, that has been devised to solve the above-described disadvantage. In FIG. 7, numerals 3, 4, 5, and 6 represent resistors. The resistors 4, 5, and 6 are connected in series between a reference voltage V.sub.ref and the output of a differential amplifier 1 in this order from the reference voltage V.sub.ref side, and the node between the resistors 4 and 5 is connected through the resistor 3 to the inverting input terminal (-) of the differential amplifier 1. An analog switch 2 is connected in parallel with the resistor 6. Note that, in the drawings and descriptions connected to the second conventional example, such elements as are found also in the first conventional example are identified with the same reference numerals and symbols, and overlapping explanations will not be repeated.
Having the construction as described above, this current-to-voltage conversion IC 30 operates as follows. When the photodiode PD that is connected to the input terminal IN receives light, it allows a current I that is proportional to the amount of the light received to flow through the feedback resistor. This raises the voltage appearing at the output terminal OUT by I.times.(R.sub.1 .times.R.sub.2 +R.sub.2 .times.R.sub.3 +R.sub.3 .times.R.sub.1)/R.sub.2 when the analog switch 2 is on and by I.times.{R.sub.1 .times.R.sub.2 +R.sub.2 .times.(R.sub.3 +R.sub.T)+(R.sub.3 +R.sub.T .times.R.sub.1 }/R.sub.2 when the analog switch 2 is off. Here, it is assumed that the resistances of the resistors 3, 4, 5, and 6 are R.sub.1, R.sub.2, R.sub.3, and R.sub.T, respectively.
Accordingly, to obtain the same conversion gain as in the current-to-voltage conversion IC 20 of the first conventional example, these resistors are set to have resistances that satisfy (R.sub.1 .times.R.sub.2 +R.sub.2 .times.R.sub.3 +R.sub.3 .times.R.sub.1)/R.sub.2 =R, if the resistor that is connected in parallel with the analog switch 2 is ignored.
On the other hand, in the current-to-voltage conversion IC 30 of the second conventional example, the resistance R.sub.F of the feedback resistor is given as R.sub.F =(R.sub.1 .times.R.sub.2 +R.sub.2 .times.R.sub.3 +R.sub.3 .times.R.sub.1)/(R.sub.2 +R.sub.3)&lt;(R.sub.1 .times.R.sub.2 +R.sub.2 .times.R.sub.3 +R.sub.3 .times.R.sub.1)/R.sub.2 =R. This means that, in the second conventional example, the same conversion gain as in the first conventional example is obtained with a lower resistance in the feedback resistor. More specifically, if it is assumed that R=12 k.OMEGA., the same conversion gain is obtained when R.sub.1 =R.sub.3 =2 k.OMEGA. and R.sub.2 =500.OMEGA.. At this time, the feedback resistor has a resistance R.sub.F =2.4 k.OMEGA., which is one fifth as large as that in the first conventional example.
As a result, in the current-to-voltage conversion IC 30 of the second conventional example, the same conversion gain as in the first conventional example is obtained with a higher second pole frequency (1/2 .pi.CR.sub.F), and thus with a wider frequency range.
In the current-to-voltage conversion IC 30 of the second conventional example, turning the analog switch 2 from on to off results in increasing the conversion gain. This is achieved by adjusting the resistance of that one of the feedback-resistor-constituting resistors which is connected to the output, i.e. the lower-impedance side, of the differential amplifier 1. Accordingly, the effect of adjusting the resistance of this resistor on the second pole, that is, the variation of the second pole frequency resulting therefrom can be reduced to negligibly small by setting the resistor 4 to have as low a resistance R.sub.2 as possible, provided that it does not make the loop gain unduly low and the frequency range unduly narrow.
Nevertheless, when the analog switch 2 is turned from on to off, the feedback ratio changes from R.sub.2 /(R.sub.2 +R.sub.3) to R.sub.2 /(R.sub.2 +R.sub.3 +R.sub.T), causing the loop gain (the gain of the feedback loop) to drop. This means that, as the conversion gain is increased, the available frequency range becomes narrower, just as in the current-to-voltage conversion IC 20 of the first conventional example.
FIG. 8 shows the frequency response of the current-to-voltage conversion IC 30. In FIG. 8, the following symbols are used:
In a CD player, it is necessary to cope with different reflectances between a CD-ROM, which is a read-only medium, and a CD-R, which is a recordable medium; in a DVD player, it is necessary to cope with different levels between the signals reproduced from the first and second layers of a disk. Thus, in these players, it is essential that the conversion gain be variable. However, as described above, in conventional current-to-voltage conversion ICs, as the conversion gain is increased, the available frequency range becomes narrower. As a result, especially in CD players that offer more than 20 times the standard speed and in DVD players that offer three times the standard speed, where RF signals typically reach frequencies as high as 30 to 50 MHz, it has been impossible, with a conventional current-to-voltage conversion IC, to play back with uniform quality a CD-ROM and a CD-R in a CD player and the first and second layers of a disk in a DVD player.